Question

In young'slit double slit experiment,the intensity at a point is 1/4 of is the maximum intensity.angular position at this point is

Answer 1

I=Imaxcos2(ϕ2)I=Imaxcos2⁡(ϕ2)∴Imax4∴Imax4=Imaxcos2ϕ2=Imaxcos2ϕ2cosϕ2=12cosϕ2=12or ϕ2=π3ϕ2=π3∴ϕ=2π3∴ϕ=2π3=(2πλ)=(2πλ).Δx.ΔxWhere Δx=dsinθΔx=dsin⁡θSubstituting in equation we get,sinθ=λ3dsin⁡θ=λ3dor θ=sin−1(λ3d)θ=sin−1⁡(λ3d)Hence b is the correct answer.

Answer 2

Answer:

The angular position at this point is 151°.

Explanation:

Given that,

The intensity at a point is 1/4 of is the maximum intensity.

$I=\dfrac{I_{0}}{4}$

We know that,

Formula of intensity in double slit interference

$I =4I_{0}cos^{2}\dfrac{\phi}{2}$.....(I)

Where, I = intensity

$I_{0}$= maximum intensity

$\phi$= angular position

Put the value of I in equation (I)

$\dfrac{I_{0}}{4} =4I_{0}cos^{2}\dfrac{\phi}{2}$

$\dfrac{1}{16}=cos^{2}\dfrac{\phi}{2}$

$\dfrac{1}{4}=cos\dfrac{\phi}{2}$

$\phi=2cos^{-1}\dfrac{1}{4}$

$\phi=151^{\circ}$

Hence, The angular position at this point is 151°.